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Oct 18, 2017 at 14:56 vote accept VSJ
Oct 16, 2017 at 12:42 answer added Bill Bradley timeline score: 3
Oct 14, 2017 at 3:24 comment added Igor Rivin Yes, that follows from Marcel's answer, using, e.g., Markoff's inequality.
Oct 14, 2017 at 2:25 comment added VSJ If $D$ is the absolute value of the largest dot product, one could seek a bound of the sort $\mathbb P(D > \epsilon_n) < \epsilon_n$ for a suitable choice of $\epsilon_n \to 0$.
Oct 14, 2017 at 0:05 comment added Igor Rivin What do you mean by "upper-bound the largest possible dot product"? All of the $X_i$ could be equal to $10^{10^{10}}.$ You can upper bound it with high probability, but that's the best you can do.
Oct 13, 2017 at 23:41 answer added Marcel timeline score: 7
Oct 13, 2017 at 22:26 history asked VSJ CC BY-SA 3.0